// 逆序二进制的状态
// 测试链接 : https://leetcode.cn/problems/reverse-bits/

// 神的思路
class Solution 
{
public:
    uint32_t reverseBits(uint32_t n) 
    {
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
		n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
		n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
		n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
		n = (n >> 16) | (n << 16);
		return n;
    }
};

// 傻白甜的思路
class Solution 
{
public:
    uint32_t reverseBits(uint32_t n) 
    {
        uint32_t ret = 0;
        for(int i = 0; i < 32 && n > 0; ++i)
        {
            ret |= (n & 1) << (31 - i);
            n >>= 1;
        }    
        return ret;
    }
};